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Show that an equation is homogeneous?

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1 Answer

Looking at (x2 + 3xy + y2)dx - x2dy = 0, note that is is already of the M(x,y)dx + N(x,y)dy.

Let v = <x,y>.

For the two functions M(v) = x2 + 3xy + y2 and N(v) = -x2, we can establish homogeneity as follows.

M(av) = (ax)2 + 3(ax)(ay) + (ay)2 = a2x2 + 3a2xy + a2y2 = a2(x2 + 3xy + y2) = a2M(v).

N(av) = -(ax)2 = -a2x2 = a2N(v).

Since M and N are the same degree, this same scalar multiplication property holds for the whole equation.

Thus M and N are homogeneous and so is your equation.